SAMPLE EXAM 2 FALL'12, SEE
PROBLEMS 1 AND 5 FOR TEST 1 EXTRA CREDIT EXAMPLES FROM
CHAPTER 3 ON RELATIVE AND CIRCULAR MOTION. PROJECTILE MOTION
IS REQUIRED. **SCROLL DOWN TO THE
BOTTOM OF THE PAGE TO SEE THE SOLUTIONS.**

Turn in the
following exercises/problems next week:
Ch. 2 Free Fall exercise/problems 35, 40, 41, 44, 48, 47, 80, 81, 84
,88, 87, 89. (*DISCUSSIONS PROVIDED SO
FAR; some problems may not have discussions but still are due.)

COMMENTS: As a rule of thumb, graphs as in #88
and #87 are especially helpful in problems with two moving
objects. This would apply to quiz 2 as well.

MORE DISCUSSIONS TO FOLLOW: CHECK BACK FREQUENTLY

44. Study Problem 89, especially part (a), discussion below in detail. Notice
differences and similarities. In particular compare the
sandbag's motion with the helicopter's after shut
off or Austin Powers' trajectory after he "steps
off" copter such that his initial velocity in free
fall, before turning on jets, is up relative to
ground. Same equations/formulae. In both problems,
Austin Powers and sandbag move upward initially relative to
ground despite "stepping off" with zero velocity with
respect to rising vessel.

80. SOLVE A VERTICAL FREE FALL EQUATION FOR TIME OF FLIGHT OF
EGG. THEN SUBSTITUTE INTO
PROFESSOR'S UNIFORM HORIZONTAL MOTION EQUATION.

84. SET UP THE SYSTEM OF EQUATIONS FOR THE PROBLEM.
TIP: LET DOWN BE POSITIVE. Thus a = +g. Set the top level
(windowsill) to be Y1 = 0 , the top of the window
at Y2 and the bottom of the window at Y3. Note: Since the pot was
dropped, V1 = 0. You are given 1.90 m = (V3 + V2)*(0.420 s) /2
and you also know
V3 = V2 + g*(0.420 s) . Solve these two equations
simultaneously for V3 and V2. Then use Y2 - Y1 =
V22/(2g).

Note: t3 - t2 = 0.420 seconds.

YOU CAN ALSO FIND THE SOLUTION USING ANOTHER SET OF EQUATIONS
RELYING MORE ON TIME.
I.E., Y3 - Y2 = 1.90 m = V2*(t3 - t2)+ (1/2)*g*(t3
- t2)2 from which you can get V2, requested
quantity.

THE PROBLEM COULD BE RELEVANT IN A CRIME OR INSURANCE
INVESTIGATION: For example investigators are trying to find
V2 to determine the speed of t pot at impact with someone
injured while walking on ground below.

88**. Let down be the positive direction of motion.
The triangular shape represents the sound's motion, tb
the time it takes sound to get to bottom, 3.0 seconds the
time it takes sound and professor to meet while she moves downward
in positive direction, and D the cliff's height. The
parabola represents her free fall motion. The
slope of inclined triangular sections are + 340 m/s and -
340 m/s moving right in time, respectively. Solve for D =
340*tb. Use information to find tb,
geometrically if necessary. 3.0 seconds is the time when the two
objects have the same X; use properties of lines,
parabolas vertexed at origin.

FURTHER COMMENTS:
Think of a horizontal one dimensional race of sorts between
a sound pulse and
someone speeding
up to the right with acceleration 9.8 m/s^2. As soon as I start I
send a sound signal in the forward direction toward a vertical wall
at the far right.. That signal will bounce off the wall and pass
through me as it moves in the opposite direction and I move in the
initial direction toward the wall. An echo. The shape of the sound
signal's position vs time will be a triangle. Like a pyramid. As
you move to the right in time, the pyramid slopes upward at an
incline whose tangent equals the speed of sound. At peak of pyramid, the sound hits the wall, and as
the pulse returns in the opposite direction, the curve traced out is a
straight line with decreasing values of D as you move away
from peak in time—the slope of this segment is the negative of
the speed of sound. Find the intersection of pyramid segment and
parabola representing your position vs time given by
(1/2(*g*t^2.

This requires some care since mathematically it could
be on either side of the peak in time. To the left of the peak in
time means we catch up with the sound, highly unlikely and
mathematically impossible since the speed of sound is over 300 m/s
and we start from rest. In theory I could catch up with the sound if
someone launched me with huge initial velocity from a strong enough spring
loaded cannon (LOL). So the intersection is to the right of the peak
in time. Use the equation of a line with
negative slope -340 m/s. The intersection between the parabola and
that line
segment happens at 3 seconds.

The value of the pyramid will be D – 340*(t – tb), where tb is the
time it takes the sound to hit the bottom, t = 3 seconds and
D is the depth of the cliff whose bottom the sound
hits. You have :

D – 340*(t – tb)= (1/2)*gt2, where t = 3
seconds. .

We need to find D and we will know everything we need to know about
the situation. But D = 340*tb . Thus,

NOTE: THE DIAGRAM ABOVE IS NOT TO SCALE AND SHOULD BE
RE-DRAWN TO BETTER REFLECT
THE SITUATION. IN PARTICULAR,
THE PICTURELOOKS
MORE LIKE THIS:

NOTE THE POOR PROFESSOR, WHO REPRESENTS THE
PARABOLA ABOVE, REACHES DEPTH D IN A TIME OVER THREE
TIMES 3.0 SECONDS, THE INSTANT HE/SHE HEARS THE
ECHO.

THE TWO DOTS REPRESENTS A CONTINUATION OF THE PARABOLA TO
THE POINT WHERE PROFESSOR HITS GROUND, the parabola reaching
value D. (HOPE A TRAMPOLINE ALLOWED HIM/HER TO
SURVIVE---MAYBE HE/SHE LANDED ON REALLY SOFT
PILLOWS!).

87. Let up be the positive direction. (a) and (b) are
basic as is (d) once you get (c):

(c)

At time t = 0, the first ball is dropping from rest downward from
height 3.0 m and second ball is thrown upward with speed equal to
two-thirds the value computed in part (a). Two
parabolas are shown , vertexed at point (0, 3.0), where 3.0
is in meters (m) and at (tm, hm),
where tm and hm are the time and value
of maximum height reached by second ball. Solve
for tf, the time when they pass.
(d) y = (2/3)*v1 - (1/2)*g*t2. , where
t is time from part (c) and is
velocity from part (a)

89. SYNOPSIS OF HINT TO #89 PART (b). NOTE: t^2 means t2. FOR
PART (a), see the methods of #44 above.

# (a) THE HELICOPTER SHUTS OFF ENGINES AT A HEIGHT Y2 OF 250 m = (1/2)(5)(10)2.
AT
THAT MOMENT THE HELICOPTER HAS VELOCITY V2 = (5)*(10) = 50 m/s.
THIS
IS ALSO THE INITIAL HEIGHT AND VELOCITY OF AUSTIN POWERS WHEN HE
STEPS
OFF THE COPTER; FROM THIS INFO YOU CAN GET THE MAXIMUM
HEIGHT REACHED BY COPTER AND POWERS, WHO ARE BOTH
IN FREE FALL AND MOVE TOGETHER BEFORE POWERS TURNS ON
HIS "JET PACKS." HE TURNS ON HIS JET PACKS AFTER
HE AND COPTER REACH MAXIMUM HEIGHT AND ARE BOTH MOVING DOWN.